1 Coulombs Law and Electric Fields Physics 102: Lecture 02 Today we will get some practice using Coulombs Law learn the concept of an Electric Field.

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Coulomb’s Lawand Electric Fields

Physics 102: Lecture 02

Today we will … • get some practice using Coulomb’s Law• learn the concept of an Electric Field

Note the change in theDaily Planner

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But first, a leftover from last lectureACT: Induced Dipole

1) Nothing2) Attracted to charged sphere.3) Repelled from charged sphere.

• An uncharged conducting sphere is hung next to a charged sphere. What happens when the uncharged sphere is released?

Van de Graaff demo

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Recall Coulomb’s Law• Magnitude of the force between charges q1 and q2

separated a distance r:F = k q1q2/r2 k = 9x109 Nm2/C2

• Force is – attractive if q1 and q2 have opposite sign

– repulsive if q1 and q2 have same sign

• Units:– q’s have units of Coulombs (C)

• charge on proton is 1.6 x 10-19 C– r has units of m– F has units of N

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Three Charges

Q=-7.0 CQ=+7.0C

Q=+2.0C

6 m

4 m

• Calculate force on +2C charge due to other two charges– Calculate force from +7C charge– Calculate force from –7C charge– Add (VECTORS!)

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2

3

4

5

6

7

8

9

10

11

12

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F7

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Three Charges

Q=-7.0 CQ=+7.0C

Q=+2.0C

6 m

4 m

• Calculate force on +2C charge due to other two charges– Calculate force from +7C charge– Calculate force from –7C charge– Add (VECTORS!)

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F+7

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Q=-7.0 CQ=+7.0C

Q=+2.0C

6 m

4 m

• Calculate force on +2C charge due to other two charges– Calculate force from +7C charge– Calculate force from –7C charge– Add (VECTORS!)

F+7

N25

)107)(102)(109( 669

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F

N105 37

F9 6 6

7(9 10 )(2 10 )(7 10 ) N

25F

37 5 10 NF

5 m

• F = k q1q2/r2

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Three Charges

F-7

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Q=-7.0 CQ=+7.0C

Q=+2.0C

6 m

4 m

F+7

5 m14

F-7

Adding Vectors F+7+F-7

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Q=-7.0 CQ=+7.0C

Q=+2.0C

6 m

4 m

F+7

5 m14

F-7

Adding Vectors F+7+F-7

y components cancel

x components:

F = F+7,x + F-7,x

F+7,x = (3/5)F+7

F-7,x = (3/5)F-7

F = (3/5)(5+5)x10-3 N=6 x 10-3 N

F

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Electric Field• Charged particles create electric fields.

– Direction is the same as for the force that a + charge would feel at that location.

– Magnitude given by: E F/q = kq/r2

+

r = 1x10-10 m

Qp=1.6x10-19 C

E

E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right)

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Preflight 2.3What is the direction of the electric field at point B?

1) Left

2) Right

3) Zero

x

yA

B

72% 16% 9%

“B only has the charge from the negative which is pushing away from itself .”

“it is closer to the negative charge, and the field lines point toward negative charges .”

“electric fields of equal magnitudes but opposite directions are present due to the positive and negative charges .”

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Since charges have equal magnitude, and point B is closer to the negative charge net electric field is to the left

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ACT: E FieldWhat is the direction of the electric field at point C?

A. Left

B. Right

C. Zero

x

y

C

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Away from positive charge (right)

Towards negative charge (right)

Net E field is to right.

Red is negative

Blue is positive

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Comparison:Electric Force vs. Electric Field

• Electric Force (F) - the actual force felt by a charge at some location.

• Electric Field (E) - found for a location only – tells what the electric force would be if a charge were located there:

F = Eq• Both are vectors, with magnitude and direction.

Add x & y components. Direction determines sign.

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Preflight 2.2What is the direction of the electric field at point A?

1) Up

2) Down

3) Left

4) Right

5) Zerox

yA

B

6%8%3%58%25%

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Red is negative

Blue is positive

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ACT: E Field IIWhat is the direction of the electric field at point A, if

the two positive charges have equal magnitude?

A. Up

B. Down

C. Right

D. Left

E. Zero

x

yA

B

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Red is negative

Blue is positive

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Electric Field of a Point Charge

+E

2.91011 N/C

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1011

N/C 0.8

1011 N/C

2510 11 N/C

This is becoming a mess!!!40

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Electric Field Lines• Closeness of lines shows

field strength- lines never cross

• # lines at surface Q

• Arrow gives direction of E- Start on +, end on -

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A B

X

Y

Preflight 2.5

Charge A is

1) positive 2) negative 3) unknown

Field lines start on positive charge, end on negative.

4493% 4% 3%

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A B

X

Y

Preflight 2.6

Compare the ratio of charges QA/ QB

1) QA= 0.5QB 2) QA= QB 3) QA= 2 QB

# lines proportional to Q

12% 9% 63% 45

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A B

X

Y

Preflight 2.8

The magnitude of the electric field at point X is greater than at point Y

1) True 2) False Density of field lines gives E46 14%

86%

The electric field is stronger when the lines are located closer to one another.

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ACT: E Field Lines

Compare the magnitude of the electric field at point A and B

1) EA>EB 2) EA=EB 3) EA<EB

A

B

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E inside of conductor• Conductor electrons free to move

– Electrons feels electric force - will move until they feel no more force (F=0)

– F=Eq: if F=0 then E=0• E=0 inside a conductor (Always!)

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A B

X

Y

Preflight 2.10

"Charge A" is actually a small, charged metal ball (a conductor). The magnitude of the electric field inside the ball is:

(1) Negative (2) Zero (3) Positive

50 9% 74% 18%

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Recap• E Field has magnitude and direction:

– EF/q– Calculate just like Coulomb’s law– Careful when adding vectors

• Electric Field Lines– Density gives strength (# proportional to charge.)– Arrow gives direction (Start + end on -)

• Conductors– Electrons free to move E=0

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To Do• Read Sections 17.1-17.3• Extra problems from book Ch 16:

– Concepts 9-15– Problems 11, 15, 23, 27, 29, 39

• Do your preflight by 6:00 AM Wednesday.

Have a great weekend.No classes Monday.

See you next Wednesday!